Particular integral of sin(ax+b)

[iVenky]

I can prove that the particular integral of sin(ax+b) is given by

$\frac{1}{f(-a^{2})} sin(ax+b)$

That's not an issue but what happens if f(-a²) is 0 ?

I see that it is given by

$x \frac{1}{f'(-a^{2})} sin(ax+b)$

but I can't prove this though I know to prove this for e^axThanks a lot :)

 

[HallsofIvy]

I don't understand what you are saying. The general anti-derivative of sin(ax+ b) is -(1/a) cos(ax+ b)+ C for any constant C. Yes, you can take C= 0 but where did "f" come from?

 

[iVenky]

$f(D)=(D^{n}+k_{1}D^{n-1}+k_{2}D^{n-2}+...+k_{n-1}D+k_{n})$

and I mean the differential equation is given by

$(D^{n}+k_{1}D^{n-1}+k_{2}D^{n-2}+...+k_{n-1}D+k_{n})y=X<br /> \\<br /> or \\f(D)y=X<br />$

Thanks a lot :)